a) \(\frac{5x-2}{2-2x}+\frac{2x-1}{2}+\frac{x^2+x-3}{1-x}=1\)

ĐK: x≠1

<=>\(\frac{5x-2}{2\left(1-x\right)}+\frac{2x-1}{2}\frac{x^2+x-3}{1-x}=1\)

<=>\(\frac{5x-2+\left(1-x\right).\left(2x-1\right)+2\left(x^2+x-3\right)}{2\left(1-x\right)}=1\)

<=>\(\frac{5x-2+2x-1-2x^2+x+2x^2+2x-6}{2\left(1-x\right)}=1\)

<=>\(\frac{10x-9}{2\left(1-x\right)}=1\)

<=> 10x-9=2(1-x)

<=>10x-9=2-2x

<=> 10x+2x= 2+9

<=> 12x=11

<=> x= \(\frac{11}{12}\left(tm\right)\)

b) \(\frac{6x-1}{2-x}+\frac{9x+4}{x+2}=\frac{3x^2-2x+1}{x^2-4}\)

ĐK: x≠2, x≠-2

<=>\(\frac{6x-1}{-\left(x-2\right)}+\frac{9x+4}{x+2}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)

<=> -(x+2).(6x-1)+(x-2).(9x+4)-(3x²-2x+1)=0

<=> -(6x²-x+12x-2)+9x²+4x-18x-8-3x²+2x-1 = 0

<=> -6x²-11x+2+9x²+4x-18x-8-3x²+2x-1=0

<=> -23x-7=0

<=> -23x=7

<=> x= \(\frac{-7}{23}\left(tm\right)\)

tham khảo câu d trong

https://hoc24.vn/hoi-dap/question/919967.html

c) \(\frac{1}{x-1}\)+\(\frac{2x^2-5}{x^3-1}\)=\(\frac{4}{x^2+x+1}\) (ĐKXĐ:x≠1)

⇔\(\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)+\(\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}\)=\(\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

⇒x²+x+1+2x²-5=4x-4

⇔3x²-3x=0

⇔3x(x-1)=0

⇔x=0 (TMĐK) hoặc x=1 (loại)

Vậy tập nghiệm của phương trình đã cho là:S={0}

\(\frac{4}{x^2-3x+2}-\frac{3}{2x^2-6x+1}+1=0\) \(Đkxđ:.......\)

Đặt: \(t=x^2-3x+2\left(t\ne0\right)\)

\(\Rightarrow2t=2x^2-6x+4\)

\(\Rightarrow2x^2-6x+1=2t-3\)

\(Pt:\Leftrightarrow\frac{4}{7}-\frac{3}{2t-3}+1=0\)

\(\Leftrightarrow4\left(2t-3\right)-3t+t\left(2t-3\right)=0\)

\(\Leftrightarrow8t-12-3t+2t^2-3t=0\)

\(\Leftrightarrow2t^2+2t-12=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-3\end{matrix}\right.\left(tm:\left[{}\begin{matrix}t\ne0\\t\ne\frac{3}{2}\end{matrix}\right.\right)\)

+ Với \(t=2\) thì: \(x^2-3x+2=2\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\left(tmđk\right)\)

+ Với \(t=-3\) thì \(x^2-3x+2=-3\)

\(\Leftrightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}=0\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}=0\left(vô-lí\right)\)

Vậy pt có nghiệm: \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Bài 2:

ĐKXĐ: $x\neq 1;2;3;6$

PT $\Leftrightarrow \frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}-\frac{1}{x-1}$

$\Leftrightarrow \frac{5x-12}{x^2-5x+6}=\frac{5x}{x^2-7x+6}$

Đặt $x^2+6=t$ thì $\frac{5x-12}{t-5x}=\frac{5x}{t-7x}$

$\Rightarrow (5x-12)(t-7x)=5x(t-5x)$

$\Leftrightarrow 10x^2+12t+84x=0$

$\Leftrightarrow 10x^2+12(x^2+6)+84x=0$

$\Leftrightarrow 22x^2+84x+72=0$

$\Leftrightarrow 11x^2+42x+36=0$

$\Rightarrow x=\frac{-21\pm 3\sqrt{5}}{11}$

\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)

\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)  

PS: Điều kiện xác đinh bạn tự làm nhé 

từ đề\(\Leftrightarrow\frac{x-1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{2x-2}{x\left(x-2\right)-4\left(x-2\right)}+\frac{3x-3}{x\left(x+1\right)-2\left(x+1\right)}+\frac{4x-4}{x\left(x+1\right)+5\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}+\frac{3}{\left(x-2\right)\left(x+1\right)}+\frac{4}{\left(x+1\right)\left(x+5\right)}=0\right)\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{x-4}-\frac{1}{x-5}+\frac{1}{x-2}-\frac{1}{x-4}+\frac{1}{x-2}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x-5}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2}{x-2}-\frac{2}{x-5}\right)=0\) vì \(\frac{2}{x-2}-\frac{2}{x-5}\)luôn khác 0 nên x-1=0 nên x=1.

Điều kiện xác định : x khác 4,5,2,-1. Do đó x=1 thỏa mãn. Vậy x=1

À,CHỈ CÓ 1 SỐ "0" THÔI NHÉ!

\(=>\frac{8}{2x^2-6x+2}-\frac{3}{2x^2-6x+2}=-1\)

\(=>\frac{5}{2x^2-6x+2}=-1\)

\(=>2x^2-6x+2=-5\)

\(=>2x^2-6x=-7\)

\(=>x.\left(2x-6\right)=-7\)

\(=>2x-6=-\frac{7}{x}\)

\(=>2x=\frac{-7+6x}{x}\)

\(=>3x=-7+6x\)

\(=>-7=-3x\)

\(=>x=\frac{-7}{-3}=\frac{7}{3}\)

E ms lớp 7 nên giải hơi dài thông cảm ạ :>

Huh?Sao cái  \(\frac{4}{x^2-3x+2}\)nhân cả tử và mẫu với 2 thành ra \(\frac{8}{2x^2-6x+2}\)zậy?Tui tưởng là bằng \(\frac{8}{2x^2-6x+4}\)chứ?

Đề bài yêu cầu gì bạn?